Cái này y hệt cái đề mik thi:)

trời dài thế làm lâu phết đó nha hừm làm theo đúng công thức là được :)

b) 24x^2+6x^2y−2x−12yx−3y^2x+y 

tôi làm theo cách tìm tích số

nếu thấy đúng thì tick cho tôi nha

1) Ta có: \(x^2-4xy+4y^2\)

\(=x^2-2.x.2y+\left(2y\right)^2\)

\(=\left(x-2y\right)^2\)

Phép tính trở thành: \(\left(x-2y\right)^2:\left(x-2y\right)=x-2y\)

2) Ta có: \(25x^2+2xy+\dfrac{1}{25}y^2\)

\(=\left(5x\right)^2+2.5x.\dfrac{1}{5}y+\left(\dfrac{1}{5}y\right)^2\)

\(=\left(5x+\dfrac{1}{5}y\right)^2\)

Phép tính trở thành: \(\left(5x+\dfrac{1}{5}y\right)^2:\left(5x+\dfrac{1}{5}y\right)=5x+\dfrac{1}{5}y\)

1) (x² - 4xy + 4y²) : (x - 2y)

= (x - 2y)² : (x - 2y)

= x - 2y

2) (25x² + 2xy + 1/25 y²) : (5x + 1/5 y)

= 5x + 1/5 y)² : (5x + 1/5 y)

= 5x + 1/5 y

a: \(\dfrac{1}{2}x^2\cdot2x^3-4x^2+3=x^5-4x^2+3\)

b: \(2y\left(xy-1\right)\left(xy+1\right)=2y\left(x^2y^2-1\right)=2x^2y^3-2y\)

h: \(=\left(x+3\right)\cdot\left(x^2-3x+9\right)-4x\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2-7x+9\right)\)

\(\left(x+2y\right)\left(x^2-2xy+4y^2\right)=0\)

\(\Leftrightarrow x^3+8y^3=0\)

\(\Leftrightarrow x^3=-8y^3\)

\(\left(x-2y\right)\left(x^2+2xy+4y^2\right)=16\)

\(\Leftrightarrow x^3-8y^3=16\)

\(\Leftrightarrow-8y^3-8y^3=16\)

\(\Leftrightarrow y^3=-1\Rightarrow y=-1\Rightarrow x=2\)

`-1/3x^5y^2:(-2xy)-(x^2+2x+1):(x+1)`

`=-1/3:(-2).(x^5:x).(y^2:y)-(x+1)^2:(x+1)`

`=-1/6x^4y-(x+1)`

`=-1/6x^4y-x-1`

\(\dfrac{-1}{3}x^5y^2:\left(-2xy\right)-\left(x^2+2x+1\right):\left(x+1\right)\)

\(=\dfrac{1}{6}x^4y-x-1\)

a: \(=\left(3-x\right)\left(x+1\right)\)

b: \(=3x\left(x-y\right)-5\left(x-y\right)\)

=(x-y)(3x-5)

c: \(=x\left(x-y\right)-10\left(x-y\right)\)

\(=\left(x-y\right)\left(x-10\right)\)

a) \(=x\left(3-x\right)+\left(3-x\right)=\left(3-x\right)\left(x+3\right)\)

b) \(=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)

c) \(=x\left(x-y\right)-10\left(x-y\right)=\left(x-y\right)\left(x-10\right)\)

d) \(=\left(x+y\right)^2-16=\left(x+y-4\right)\left(x+y+4\right)\)

e) \(=\left(x-y\right)\left(x+y\right)-4\left(x+y\right)=\left(x+y\right)\left(x-y-4\right)\)

f) \(=9-\left(4x^2-4xy+y^2\right)=9-\left(2x-y\right)^2=\left(3-2x+y\right)\left(3+2x-y\right)\)

g) \(=y\left(y^2-2xy+x^2-y\right)\)

h) \(=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)

i) \(=x\left(x-y\right)+\left(x-y\right)\left(x+y\right)=\left(x-y\right)\left(2x+y\right)\)

Hi

đề khó đấy

Câu 1 : sửa x^3 - 8y^3 

chọn C 

Câu 2 : B 

Câu 1.

\(x^2-8y^3=\left(x-2y\right)\left(x^2+2y+4y^2\right)\)

Câu 2.

\(-x^2\left(3-2x\right)=-3x^2+2x^3=2x^3-3x^2\)